Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3771    Accepted Submission(s): 2586

Problem Description

As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.

To make the problem easier, I promise that B will be smaller than 100000.

Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.

 

 

Input

The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.

 

 

Output

For each test case, you have to ouput the result of A mod B.

 

 

 

Sample Input

2 3 12 7 152455856554521 3250

 

 

Sample Output

2 5 1521

 

1 //原理： (a+b)%c = ((a%c)+(b%c))%c 2   //    (a*b)%c = ((a%c)*(b%c))%c 3 #include 
        
          4 #include 
         
           5 #include 
          
            6 using namespace std; 7 #define maxn 1000+5 8  9 int num[250];10 char numc[maxn];11 12 int main()13 {14     int b;15     while(cin>>numc>>b)16     {17         int i = strlen(numc) - 1;18         int p = 1,ans = 0;19         while(i>=0)20         {21             ans = (ans + (numc[i]-'0')*p)%b;22             p *= 10;23             p %= b;24             i--;25         }26         cout<
           
            <